

A306521


Least integer N > 2 such that the number of primes (<=N) <= the number of basenzero containing numbers (<=N).


4



3, 3, 4, 28, 42, 104, 136, 329, 510, 856, 1449, 2212, 2782, 3434, 4188, 5042, 6001, 7082, 8276, 9604, 11062, 12666, 14405, 31651, 35694, 40061, 66427, 73966, 108764, 149756, 197516, 288280, 354924, 515538, 701002, 963687, 1318399, 1840377
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OFFSET

2,1


COMMENTS

a(n) <= A306526(n), equality holds for n = 2, 14, 15, 16, 17, 18, 19, 20, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52 (but a(n) < A306526(n) for all other indices n up to 82). For sufficiently large n, equality holds true for those bases n which satisfy 1/2 <= fract(sqrt(n/log(n)) + O(sqrt(log(n)/n))) < 3/4. This is true for infinitely many further indices, at least for all bases n = ceiling(x), where x is a solution of x/log(x) = kth triangular number + 1/4, k > 1. For k = 2..10 the corresponding bases are n = 19, 48, 92, 152, 230, 326, 440, 574, 727. Let e(n) be the number of bases m <= n for which a(m) = A306526(m), then lim_{n>infinity} e(n)/n >= 1/4. Conjecture: lim_{n>infinity} e(n)/n = 1/4.


LINKS

Hieronymus Fischer, Table of n, a(n) for n = 2..100


FORMULA

With numOfZeroNum_n(k) [= the number of basenzero containing numbers <= k] and pi(k) [= the number of primes <= k] and d := log(n1)/log(n):
a(n) = min(k > 2  pi(k) <= numOfZeroNum_n(k)). Because of d = d(n) < 1, numOfZeroNum_n(k) = k*(1 + O(k^(d1)), pi(k) = k/log(k)*(1+o(1)), and pi(3) = 2 >= 2 = numOfZeroNum_n(3), this minimum always exists (for n > 2). The case n = 2 is obvious. See A324160 regarding general formulas for numOfZeroNum_n(k).
Estimate of the nth term:
a(n) > e*(1 + c1/c2*(1 + sqrt(1 + c2*c3/c1^2)))^(1/(1d)), for n > 6,
where d := log(n1)/log(n),
c0 := e^(1d),
c1 := (n1)^d/(n2)  1/e^(sqrt(n*log(n)))  d*c0,
c2 := (n1)^d/(n2)  1/e^(sqrt(n*log(n))) + (1  1/sqrt(n*log(n)))*c0,
c3 := 2*(1d)*c0.
Also, but less accurate, for n > 6,
a(n) > e*(1 + 1/(sqrt(n*log(n))  2))^(1/(1d)).
a(n) > e*(1 + 1/(sqrt(n*log(n))  2))^((n1/2)*log(n)).
a(n) <= A306526(n), see A306526 for further upper bound estimations.
Asymptotic behavior:
a(n) = O(sqrt(n)*e^sqrt(n*log(n))).
lim sup a(n)/e^(sqrt(n*log(n))+(log(n)+1)/2) = 1, for n > infinity.
lim inf a(n)/e^(sqrt(n*log(n))+1/2) = 1, for n > infinity.


EXAMPLE

a(2) = 3, since pi(3) = 2 <= 2 = numOfZeroNum_2(3), where numOfZeroNum_2(m) is the number of base2zero containing numbers <= m and pi(m) = number of primes <= m. The first base2zero containing numbers are 0 = 0_2, 2 = 10_2, 4 = 100_2, ... (Hint: numbers <= 2 are out of scope for selfevident reasons).
a(3) = 3, since pi(3) = 2 <= 2 = numOfZeroNum_3(3), where numOfZeroNum_3(m) is the number of base3zero containing numbers <= m and pi(m) = number of primes <= m. The first base3zero containing numbers are 0 = 0_2, 3 = 10_3, 6 = 20_3, 9 = 100_3, 10 = 101_3, 11 = 102_3, 12 = 120_3, ...
a(4) = 4, since pi(3) = 2 > 1 = numOfZeroNum_4(3), pi(4) = 2 <= 2 = numOfZeroNum_4(4), where numOfZeroNum_4(m) is the number of base4zero containing numbers <= m and pi(m) = number of primes <= m. The first base4zero containing numbers are 0 = 0_2, 4 = 10_4, 8 = 20_4, ...


PROG

(PARI) cz(m, n) = vecmin(digits(m, n))==0;
a(n) = {my(m=2, nbz=1+sum(k=1, 2, cz(k, n)), pmp=primepi(2)); for (m=3, oo, if (isprime(m), pmp++); if (cz(m, n), nbz++); if (pmp <= nbz, return (m)); ); } \\ Michel Marcus, Jun 10 2019


CROSSREFS

Cf. A011540, A052382, A306195, A306442, A306526, A324154, A324155, A324160, A324161.
Sequence in context: A118114 A266067 A300368 * A300369 A304296 A076183
Adjacent sequences: A306518 A306519 A306520 * A306522 A306523 A306524


KEYWORD

nonn,base


AUTHOR

Hieronymus Fischer, Mar 29 2019


STATUS

approved



